OFFSET
0,2
COMMENTS
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A289780 for a guide to related sequences.
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9, -27, 55, -36, 55, -27, 9, -1)
FORMULA
G.f.: (1 + x - x^2 + 34 x^3 - x^4 + x^5 + x^6)/(1 - 9 x + 27 x^2 - 55 x^3 + 36 x^4 - 55 x^5 + 27 x^6 - 9 x^7 + x^8).
a(n) = 9*a(n-1) - 27*a(n-2) + 55*a(n-3) - 36*a(n-4) + 55*a(n-5) - 27*a(n-6) + 9*a(n-7) - a(n-8).
MATHEMATICA
z = 60; s = x*(1 + 4*x + x^2)/(1 - x)^4; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000578 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289799 *)
LinearRecurrence[{9, -27, 55, -36, 55, -27, 9, -1}, {1, 10, 62, 377, 2232, 13015, 75898, 444014}, 30] (* Harvey P. Dale, Jan 07 2024 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 12 2017
STATUS
approved